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Is it true that for all topological spaces $X$ and all open $A,B \subseteq X$, it holds that if $A \neq B$, then $\partial A \neq \partial B$? What about if $A$ and $B$ are instead assumed closed? I can't think of any counterexamples, but my gut feeling is that for open sets its true but for closed sets it may fail.

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You can even have three disjoint connected open subsets of the plane with same boundary, the Lakes of Wada. –  Stefan Hamcke Mar 2 '13 at 13:45
    
@StefanH. Woah! Cool counterexample. –  goblin Mar 2 '13 at 21:37
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up vote 6 down vote accepted

Let $X$ be a disk, let $A$ be an open disk inside $X$, let $B$ be the complement in $X$ of the closure of $A$. Don't $A$ and $B$ have the same boundary?

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Take $X = (0,2) \subset \mathbb{R}$, $A = (0,1), B = (1,2)$ but $\partial A = \partial B = \{ 1\}$. And $[0,1]$ and $\{0,1\}$ (in $\mathbb{R}$) are both closed and have the same boundary $\{0,1\}$ as well. So there are quite a few counterexamples, also for closed and open sets.

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